Contour integral around circle - Complex Analysis

Question

Exercise :

Show that :

$$\frac{1}{2πi}\oint_{|z|=r} \frac{e^{1/z}}{z-1}dz = \Bigg\{\begin{matrix} 1-e & r<1\\ 1 & r >1 \end{matrix} $$

I came across this exercise while I was studying for my complex analysis semester exams and I would really appreciate any guidance, tips or thorough solution. I know I am not providing an attempt (probably the only time I have done so in all my questions) but I can't grasp on how to work depending on the radius $R$. Most integrals I've solved can be worked around with the Theorem of Residues or the Trigonometric/General Integrals techniques, but I cannot seem to work on this one. Maybe one idea is to work through poles and singular points.

I would really value a thorough explanation.

Answer 1

An idea for the residues

$$z=0:\;\;\frac{e^{1/z}}{z-1}=-\frac1{1-z}\cdot e^{1/z}=-\sum_{n=0}^\infty z^n\cdot\sum_{n=0}^\infty\frac1{n!z^n}=$$

$$=-\sum_{n=0}^\infty\sum_{k=0}^n\frac{z^k}{(n-k)!z^{n-k}}=-\sum_{n=0}^\infty\sum_{k=0}^n\frac1{(n-k)!z^{n-2k}}$$

The coefficient of $\;z^{-1}\;$ in the above is given by $\;n-2k=1\iff k=\cfrac{n-1}2\;$ , and thus $\;n\;$ must be odd, so we get that the coefficient for each such value of $\;n\;$ odd is $\;\frac1{\left(\frac{n+1}2\right)!}\;$ , and thus the whole wanted coefficient is

$$-\sum_{n=1}^\infty\frac1{n!}=-(e-1)=1-e$$

I think the residue at $\;z=1\;$ is much easier to calculate and I leave that to you.

Answer 2

Hint

• If $r<1$, the only singularity is at $0$. Since $$\frac{1}{z-1}=-\sum_{n\geq 0}z^n,$$ you can conclude easily compute the Laurent series of $\frac{e^{1/z}}{z-1}$ and conclude.

• If $r>1$, you also have a singularity at $z=1$. Since $e^{1/z}$ is holomorphic at the neighborhood of $1$, you can easily compute the residue at $z=1$ using Cauchy formula.